Question #57287

Two cylinders of gas are identical in every way, except one contains oxygen gas, the other contains helium gas. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater temperature increase, or will they both show the same temperature increase? Thanks.
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Expert's answer

2016-01-19T08:28:41-0500

Answer on Question #57287 Physics / Molecular Physics | Thermodynamics

Two cylinders of gas are identical in every way, except one contains oxygen gas, the other contains helium gas. Both hold the same volume of gas at STP and are closed by a movable piston at one end. Both gases are now compressed adiabatically to one-third their original volume. Which gas will show the greater temperature increase, or will they both show the same temperature increase?

Given:

i1=5i_1 = 5 - degree of freedom of two atomic oxygen (O2)(O_2)

i2=3i_2 = 3 - degree of freedom of one atomic helium (He)(He)

V0,P0,T0V_0, P_0, T_0 - initial state of each gas

V1=V2=V03V_1 = V_2 = \frac{V_0}{3} - volumes after compression

Adiabatically (Q=0)(Q = 0)

ΔT2ΔT1?\frac{\Delta T_2}{\Delta T_1} - ?

Solution:

Adiabatic process can be described by the formulas:


PVγ=const, or TVγ1=const, where γ=i+2iPV^{\gamma} = \text{const}, \text{ or } TV^{\gamma-1} = \text{const}, \text{ where } \gamma = \frac{i+2}{i}


It's convenient to use the second one in our problem. We apply this formula to the states before and after compression of each gas respectively.


T0V0γ11=T1V1γ11T_0 V_0^{\gamma_1 - 1} = T_1 V_1^{\gamma_1 - 1}T0V0γ21=T2V2γ21T_0 V_0^{\gamma_2 - 1} = T_2 V_2^{\gamma_2 - 1}γ1=75;γ2=53\gamma_1 = \frac{7}{5}; \quad \gamma_2 = \frac{5}{3}


From where


T1=T0(V0V1)γ11=T03γ11T_1 = T_0 \left(\frac{V_0}{V_1}\right)^{\gamma_1 - 1} = T_0 3^{\gamma_1 - 1}ΔT1=T0(3γ111)\Delta T_1 = T_0 (3^{\gamma_1 - 1} - 1)ΔT2=T0(3γ211)\Delta T_2 = T_0 (3^{\gamma_2 - 1} - 1)ΔT2ΔT1=3γ2113γ111=32/3132/511.96\frac{\Delta T_2}{\Delta T_1} = \frac{3^{\gamma_2 - 1} - 1}{3^{\gamma_1 - 1} - 1} = \frac{3^{2/3} - 1}{3^{2/5} - 1} \approx 1.96

Answer:

ΔT2=1.96ΔT1\Delta T_2 = 1.96 \Delta T_1. Increase in temperature of helium is 1.96 times larger than that of oxygen gas.

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